$ B = \dfrac{x-1}{2x+1}$
$\to C = A : B= \dfrac{x-1}{x^2}: \dfrac{x-1}{2x+1} = \dfrac{x-1}{x^2}. \dfrac{2x+1}{x-1} = \dfrac{2x+1}{x^2}$
$ C \ge -1$
$\to \dfrac{2x+1}{x^2} \ge -1 $
$\to \dfrac{-2x-1}{x^2} \le 1 $
$\to -2x -1 \le x^2$
$\to x^2 +2x +1 \ge 0$
$\to (x+1)^2 \ge 0$ ( luôn đúng )
Vậy $ C \ge -1$